# Is this a closed spacelike curve?

• I
• space-time
In summary, according to the notes that I have been reading, a curve on a Lorentzian manifold can be classified as follows:-If you have a parameterized curve ξ(s) (where s is the parameter to the curve ξ) on some interval (s1, s2), then ξ is timelike if gabξ'aξ'b < 0 for all s (just so you know, those are derivatives of the curve w/ respect to s)-ξ is spacelike if gabξ'aξ'b > 0 for all s
space-time
I've been refurbishing my understanding of some relativistic concepts and I've been specifically studying the concepts of spacelike, timelike and lightlike curves. According to the notes that I have been reading, curves on a Lorentzian manifold can be classified as follows:

If you have a parameterized curve ξ(s) (where s is the parameter to the curve ξ) on some interval (s1, s2), then...

ξ is timelike if gabξ'aξ'b < 0 for all s (just so you know, those are derivatives of the curve w/ respect to s)
ξ is spacelike if gabξ'aξ'b > 0 for all s

Now, I personally wanted to see an example of a timelike curve (specifically a closed timelike curve), and I know that the Godel metric has closed timelike curves. That is why I did some calculations involving the Godel metric. Now here is what I did:

Firstly, here is the spacetime interval for the Godel metric:
ds2 = (1/2ω2)[ -(cdt + exdz)2 + dx2 + dy2 + (1/2)e2xdz2 ]

Now here are the non-zero metric tensor components:
g00 = (-1/2ω2)
g03 = g30 = (-ex/2ω2)
g11 = g22 = (1/2ω2)
g33 = (-e2x/4ω2)

Every other element is 0.

Now here is the curve ξ(s) that I defined and parameterized:

ξ is the top face of a cylinder (standing upright). The top face is at height z = 10 and the radius of the cylinder is 4. The top face of the cylinder is traced counter-clockwise over the interval (0 , 2π) (which basically means that our curve is simply a circle of radius 4). Having said all of this, here is the parameterization:

ξa(s) = [ct , 4cos(s), 4sin(s), 10] (I did not specify a t value because the t value shouldn't matter here as far as I can tell since t does not depend on s. Technically the z value shouldn't matter either since it does not depend on s).

It follows then that the derivative of this vector with respect to s is:

ξ'a(s) = [0 , -4sin(s), 4cos(s), 0]

Now then, when I do the summation of gabξ'aξ'b , I get:

(16 / 2ω2)[sin2(s) + cos2(s)] = 8 / ω2

That final value of 8 / ω2 is positive for all s.

Now according to what I stated earlier, this would make my curve a space-like curve (not a time-like curve). Furthermore, the circular face of a cylinder is definitely a closed curve if your interval is (0 , 2π).

It would seem to me then, that I have specified a closed space-like curve. I was expecting (hoping for) a closed time-like curve, and I had read that the Godel metric apparently has CTC's running through every event.

That is why I want to ask and verify:

Have I come to the correct understanding with the work that I have done? Specifically, have I accurately specified an example of a closed space-like curve in a Godel spacetime, or did I do something wrong? Is my parameterization wrong? Is my curve just a "bad curve" or something like that? Did I simply make an arithmetic error that anyone sees?

I would appreciate any assistance. Thank you.

space-time said:
Now according to what I stated earlier, this would make my curve a space-like curve (not a time-like curve). Furthermore, the circular face of a cylinder is definitely a closed curve if your interval is (0 , 2π).

It would seem to me then, that I have specified a closed space-like curve. I was expecting (hoping for) a closed time-like curve, and I had read that the Godel metric apparently has CTC's running through every event.
That CTCs exist does not mean all closed curves are time-like.

space-time said:
ξ is timelike if gabξ'aξ'b < 0 for all s (just so you know, those are derivatives of the curve w/ respect to s)
ξ is spacelike if gabξ'aξ'b > 0 for all s
One warning: this depends on your choice of sign convention. If you diagonalise your metric you will find that the signs along the diagonal are -+++, but the opposite sign convention also appears. You are correct in this case, but do check the convention of any source you read (painful experience speaking).

Yes, the curve you've specified is a closed spacelike curve. This is because it is the perimeter of a 2d object specified at some constant coordinate time ##t##. Your time coordinate wouldn't be much of a time coordinate if constant time included unambiguous changes in time.

The point about a closed timelike curve is that it is everywhere timelike. So your curve isn't an appropriate candidate. As Orodruin says, that all points have a CTC through them does not mean that all curves are CTCs. You need to start with a timelike vector, at least.

Ibix said:
One warning: this depends on your choice of sign convention. If you diagonalise your metric you will find that the signs along the diagonal are -+++, but the opposite sign convention also appears. You are correct in this case, but do check the convention of any source you read (painful experience speaking).

Yes, the curve you've specified is a closed spacelike curve. This is because it is the perimeter of a 2d object specified at some constant coordinate time ##t##. Your time coordinate wouldn't be much of a time coordinate if constant time included unambiguous changes in time.

The point about a closed timelike curve is that it is everywhere timelike. So your curve isn't an appropriate candidate. As Orodruin says, that all points have a CTC through them does not mean that all curves are CTCs. You need to start with a timelike vector, at least.

Just to make sure of something, from what I have gathered, a vector Va is timelike if
gabVaVb < 0 (assuming the - + + + signature). For a timelike vector, if
V0 < 0 then the vector is past oriented and if V0> 0 then it is future oriented.
Are these statements correct?

space-time said:

Just to make sure of something, from what I have gathered, a vector Va is timelike if
gabVaVb < 0 (assuming the - + + + signature). For a timelike vector, if
V0 < 0 then the vector is past oriented and if V0> 0 then it is future oriented.
Are these statements correct?
The first part is true. The second isn't, except in the sense of being true for the most common coordinate choices. A simple counterexample is the the Schwarzschild black hole interior in standard Schwarzschild coordinates. In this case r is the timelike coordinate and decreasing r direction represents the future, while increasing is the past direction.

More generally, given a manifold with a pseudo-Riemannian metric, it may or may not be possible to assign a globally consistent time orientation. If it is, the manifold is said to be orientable. However, the choice of future versus past in an orientable manifold is just that - a choice. It just changes the physical interpretation of the manifold. Thus, in the example above, the choice of which r direction is future determines whether one is describing a white hole interior versus a black hole interior. Treat increasing r as future, and you have a white hole interior. Treat decreasing r as future, and you have black hole interior.

PAllen said:
More generally, given a manifold with a pseudo-Riemannian metric, it may or may not be possible to assign a globally consistent time orientation. If it is, the manifold is said to be orientable.
Time orientable to be more precise. There are also similar considerations for space.

PAllen

## 1. What is a closed spacelike curve?

A closed spacelike curve is a type of curve in spacetime that is closed, meaning it starts and ends at the same point, and is composed of entirely spacelike vectors. This means that the curve does not intersect itself and is always perpendicular to the time axis.

## 2. How is a closed spacelike curve different from a closed timelike curve?

A closed spacelike curve is different from a closed timelike curve in that it is composed of spacelike vectors, while a closed timelike curve is composed of timelike vectors. This means that a closed spacelike curve is always perpendicular to the time axis, while a closed timelike curve is always parallel to the time axis.

## 3. Is a closed spacelike curve possible in our universe?

According to current theories and observations, a closed spacelike curve is not possible in our universe. This is because it would require faster-than-light travel, which is not allowed by the laws of physics. However, some theories, such as those involving wormholes, suggest that closed spacelike curves may be possible in certain circumstances.

## 4. What are the implications of a closed spacelike curve?

If a closed spacelike curve were possible, it would have significant implications for our understanding of causality and time travel. It would also challenge our current understanding of the laws of physics, as it would require the violation of the speed of light and the principle of causality.

## 5. How do scientists study closed spacelike curves?

Scientists study closed spacelike curves through theoretical models and simulations. They also look for potential evidence of closed spacelike curves in observations of astrophysical phenomena, such as black holes and gravitational waves. However, since closed spacelike curves are not currently believed to exist in our universe, their study is largely theoretical.

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